Two adjacent sides of a parallelogram ABCD are given by \overrightarrow AB = 2\widehat i + 10\widehat j + 11\widehat k and \,\overrightarrow AD = -\widehat i + 2\widehat j + 2\widehat k The side AD is rotated by an acute angle \alpha in the plane of the parallelogram so that AD becomes AD'. If AD' makes a right angle with the side AB, then the cosine of the angle \alpha is given by

Vector Algebra PYQ — JEE Advanced 2010 · Shift 2 · Q38 with Solution

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Correct answer: B

Let $\vec{AB}=\mathbf{b}=2\hat i+10\hat j+11\hat k,$ $\vec{AD}=\mathbf{d}=-\hat i+2\hat j+2\hat k.$

We are told that $\vec{AD}$ is rotated in the plane of the parallelogram by an acute angle $\alpha$ to become $\vec{AD'}$ , and after rotation it becomes perpendicular to $\vec{AB}$ .

So, in the plane spanned by $\mathbf{b}$ and $\mathbf{d}$ , the vector $\mathbf{d}$ is turned to a new direction that is at right angle to $\mathbf{b}$ .

First find the angle $\theta$ between $\mathbf{d}$ and $\mathbf{b}$ .

Using dot product, $\mathbf{b}\cdot \mathbf{d}=(2)(-1)+(10)(2)+(11)(2)=-2+20+22=40.$

Now, $|\mathbf{b}|=\sqrt{2^2+10^2+11^2}=\sqrt{4+100+121}=\sqrt{225}=15,$ $|\mathbf{d}|=\sqrt{(-1)^2+2^2+2^2}=\sqrt{1+4+4}=\sqrt{9}=3.$

Hence, $\cos\theta=\frac{\mathbf{b}\cdot\mathbf{d}}{|\mathbf{b}|\,|\mathbf{d}|}= rac{40}{15\cdot 3}=\frac{40}{45}=\frac{8}{9}.$

So the angle between $AB$ and $AD$ is $\theta=\cos^{-1}\left(\frac{8}{9}\right).$

Since $AD'$ is perpendicular to $AB$ , the angle between $AB$ and $AD'$ is $90^\circ$ .

Because $AD$ is rotated through an acute angle $\alpha$ in the same plane to reach the perpendicular direction, we have $\alpha=90^\circ-\theta.$

Therefore, $\cos\alpha=\cos(90^\circ-\theta)=\sin\theta.$

Compute $\sin\theta$ from $\cos\theta=\frac89$ :

=\sqrt{1-\left(\frac89\right)^2} =\sqrt{1-\frac{64}{81}} =\sqrt{\frac{17}{81}} =\frac{\sqrt{17}}{9}.$$ Thus, $$\boxed{\cos\alpha=\frac{\sqrt{17}}{9}}.$$ 5. Checking options: - A: $\frac89$ ❌ - B: $\frac{\sqrt{17}}{9}$ ✅ - C: $\frac19$ ❌ - D: $\frac{4\sqrt5}{9}$ ❌ So the correct option is **B**.

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View written solutionFree

Correct answer: B

Let $\vec{AB}=\mathbf{b}=2\hat i+10\hat j+11\hat k,$ $\vec{AD}=\mathbf{d}=-\hat i+2\hat j+2\hat k.$

We are told that $\vec{AD}$ is rotated in the plane of the parallelogram by an acute angle $\alpha$ to become $\vec{AD'}$ , and after rotation it becomes perpendicular to $\vec{AB}$ .

So, in the plane spanned by $\mathbf{b}$ and $\mathbf{d}$ , the vector $\mathbf{d}$ is turned to a new direction that is at right angle to $\mathbf{b}$ .

First find the angle $\theta$ between $\mathbf{d}$ and $\mathbf{b}$ .

Using dot product, $\mathbf{b}\cdot \mathbf{d}=(2)(-1)+(10)(2)+(11)(2)=-2+20+22=40.$

Now, $|\mathbf{b}|=\sqrt{2^2+10^2+11^2}=\sqrt{4+100+121}=\sqrt{225}=15,$ $|\mathbf{d}|=\sqrt{(-1)^2+2^2+2^2}=\sqrt{1+4+4}=\sqrt{9}=3.$

Hence, $\cos\theta=\frac{\mathbf{b}\cdot\mathbf{d}}{|\mathbf{b}|\,|\mathbf{d}|}= rac{40}{15\cdot 3}=\frac{40}{45}=\frac{8}{9}.$

So the angle between $AB$ and $AD$ is $\theta=\cos^{-1}\left(\frac{8}{9}\right).$

Since $AD'$ is perpendicular to $AB$ , the angle between $AB$ and $AD'$ is $90^\circ$ .

Because $AD$ is rotated through an acute angle $\alpha$ in the same plane to reach the perpendicular direction, we have $\alpha=90^\circ-\theta.$

Therefore, $\cos\alpha=\cos(90^\circ-\theta)=\sin\theta.$

Compute $\sin\theta$ from $\cos\theta=\frac89$ :

=\sqrt{1-\left(\frac89\right)^2} =\sqrt{1-\frac{64}{81}} =\sqrt{\frac{17}{81}} =\frac{\sqrt{17}}{9}.$$ Thus, $$\boxed{\cos\alpha=\frac{\sqrt{17}}{9}}.$$ 5. Checking options: - A: $\frac89$ ❌ - B: $\frac{\sqrt{17}}{9}$ ✅ - C: $\frac19$ ❌ - D: $\frac{4\sqrt5}{9}$ ❌ So the correct option is **B**.

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